Question: Find the zeros of the function. Enter the solutions from least to greatest. $f(x) = (x + 6)^2 - 49$ $\text{lesser }x = $
Answer: $\begin{aligned} (x + 6)^2 - 49&= 0 \\\\ (x+6)^2&=49 \\\\ \sqrt{(x+6)^2}&=\sqrt{49} \end{aligned}$ $\begin{aligned} x+6&=\pm7 \\\\ x&=\pm7-6 \\ \phantom{(x + 6)^2 - 49}& \\ x=-13&\text{ or }x=1 \end{aligned}$ In conclusion, $\begin{aligned} \text{lesser }x &= -13 \\\\ \text{greater } x &= 1 \end{aligned}$